Question

What is an example of an RC circuit practice problem?

Sample practice problem -A08
A resistor `R=10Omega` is connected with `20muF and 10muF` capacitors and a `10V` DC battery as shown below. Analyze the circuit to find
1. Current in the loop after three time constants
2. Steady state energy stored in each capacitor.

Answer 1

Equivalent capacitance `C` of both capacitors connected in series is

`C_T=(C_1xxC_2)/(C_1+C_2)=(20xx10)/(20+10)=200/30=6.bar6muF`

Time constant `tau` of the circuit `RC=(10)(6.bar6xx10^-6)=6.bar6xx10^-5s`
Current `I` in the circuit is given by the expression

`I=I_0e^(-t/tau)`
where `I_0` is the initial current in the loop as the switch is closed.

`I_0=varepsilon/R=10/10=1A`

1. Current after three time constants is

   `I=I_0e^(-(3tau)/tau)`
   `=>I=1xxe^(-3)`
   `=>I=0.05A`

2. We know that
   `C_T=Q/varepsilon`
   `=>Q=6.bar6xx10^-6xx10=6.bar6xx10^-5C`

In the steady state there is no current flowing in the loop. Hence, voltage drop across resistor `=0`. Voltage of battery is divided across capacitors in the inverse fraction of respective capacitances.

`V_10=varepsilonxxC_T/C_10=10xx(6.bar6)/10=6.bar6V`

The balance voltage appears across other capacitor.

`:.V_20=10-6.bar6=3.bar3V`

Now energy stored in a capacitor is given by the expression

`E=1/2CV^2`
`:.E_10=1/2xx(10xx10^-6)(6.bar6)^2`
`=>E_10=2.bar2xx10^-4J`

Similarly

`E_20=1/2xx(20xx10^-6)(3.bar3)^2=1.bar1xx10^-4J`